Another definition for $f_n\to f$ uniformly on $E$ is that for all $\varepsilon>0$, there exists $N$ such that $n\geqslant N$ implies $|f_n(x)-f(x)|<\varepsilon$ for all $x\in E$. Note that this differs from pointwise convergence in that $N$ only depends on $\varepsilon$ and not $x$.
For example, consider the sequence of functions $f_n(x)=x^n$ on $[0,1]$. This converges to the limit function
$$ f(x)=\begin{cases}0,& x<1\\ 1,& x=1\end{cases}$$
Recall that a sequence of continuous functions that converges uniformly has a continuous limit function. Since $f$ isn't continuous, $f_n$ can't converge uniformly. In particular, $$\sup_{x\in[0,1]} |f_n(x)-f(x)|=\sup_{x\in[0,1]}x^n = 1 $$
for all $n$, so
$$ \lim_{n\to\infty}\sup_{x\in[0,1]}|f_n(x)-f(x)|\neq0. $$
In terms of the other definition, if we for example choose $\varepsilon=\frac12$, then for any $N$ we can choose $x\in(0,1)$ such that $x^N>\frac12$. Hence $f_n$ does not converge uniformly.